Mathematics » Functions II » Exponential Functions

# Functions of the Form y = ab(x+p) + q

## Functions of the form $$y=a{b}^{(x+p)}+q$$

We now consider exponential functions of the form $$y=a{b}^{(x+p)}+q$$ and the effects of parameter $$p$$.

## Optional Investigation: The effects of $$a$$, $$p$$ and $$q$$ on an exponential graph

1. On the same system of axes, plot the following graphs:

1. $$y_1 = 2^x$$
2. $$y_2 = 2^{(x – 2)}$$
3. $$y_3 = 2^{(x – 1)}$$
4. $$y_4 = 2^{(x + 1)}$$
5. $$y_5 = 2^{(x + 2)}$$

Use your sketches of the functions above to complete the following table:

 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ intercept(s) asymptote domain range effect of $$p$$
2. On the same system of axes, plot the following graphs:

1. $$y_1 = 2^{(x – 1)} + 2$$
2. $$y_2 = 3 \times 2^{(x – 1)} + 2$$
3. $$y_3 = \cfrac{1}{2} \times 2^{(x – 1)} + 2$$
4. $$y_4 = 0 \times 2^{(x – 1)} + 2$$
5. $$y_5 = -3 \times 2^{(x – 1)} + 2$$

Use your sketches of the functions above to complete the following table:

 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ intercept(s) asymptotes domain range effect of $$a$$

### The effect of the parameters on $$y = ab^{x + p} + q$$

The effect of $$p$$ is a horizontal shift because all points are moved the same distance in the same direction (the entire graph slides to the left or to the right).

• For $$p>0$$, the graph is shifted to the left by $$p$$ units.

• For $$p<0$$, the graph is shifted to the right by $$p$$ units.

The effect of $$q$$ is a vertical shift. The value of $$q$$ also affects the horizontal asymptotes, the line $$y = q$$.

The value of $$a$$ affects the shape of the graph and its position relative to the horizontal asymptote.

• For $$a>0$$, the graph lies above the horizontal asymptote, $$y = q$$.

• For $$a<0$$, the graph lies below the horizontal asymptote, $$y = q$$.

 $$p>0$$ $$p<0$$ $$a<0$$ $$a>0$$ $$a<0$$ $$a>0$$ $$q>0$$ $$q<0$$

### Discovering the characteristics

For functions of the general form: $$f(x) = y = ab^{(x+p)} + q$$:

### Domain and range

The domain is $$\{x:x\in ℝ\}$$ because there is no value of $$x$$ for which $$f(x)$$ is undefined.

The range of $$f(x)$$ depends on whether the value for $$a$$ is positive or negative.

If $$a>0$$ we have: \begin{align*} {b}^{(x+p)} & > 0 \\ a {b}^{(x+p)} & > 0 \\ a {b}^{(x+p)} + q & > q \\ f(x) & > q\end{align*} The range is therefore $$\{ y: y > q, y \in \mathbb{R} \}$$.

Similarly, if $$a < 0$$, the range is $$\{ y: y < q, y \in \mathbb{R} \}$$.

## Example

### Question

State the domain and range for $$g(x) = 5 \times 3^{(x+1)} – 1$$.

### Determine the domain

The domain is $$\{x: x \in \mathbb{R} \}$$ because there is no value of $$x$$ for which $$g(x)$$ is undefined.

### Determine the range

The range of $$g(x)$$ can be calculated from: \begin{align*} 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} – 1 & > -1\\ \therefore g(x) & > -1 \end{align*} Therefore the range is $$\{g(x): g(x) > -1 \}$$ or in interval notation $$(-1; \infty)$$.

### Intercepts

The $$y$$-intercept:

To calculate the $$y$$-intercept we let $$x=0$$. For example, the $$y$$-intercept of $$g(x) = 3 \times 2^{(x + 1)} + 2$$ is determined by setting $$x=0$$: \begin{align*} g(0) &= 3 \times 2^{(0 + 1)} + 2 \\ &= 3 \times 2 + 2\\ &= 8 \end{align*} This gives the point $$(0;8)$$.

The $$x$$-intercept:

To calculate the $$x$$-intercept we let $$y=0$$. For example, the $$x$$-intercept of $$g(x) = 3 \times 2^{(x + 1)} + 2$$ is determined by setting $$y=0$$: \begin{align*} 0 &= 3 \times 2^{(x + 1)} + 2 \\ -2 &= 3 \times 2^{(x + 1)} \\ -\cfrac{2}{3} &= 2^{(x + 1)} \end{align*} which has no real solutions. Therefore, the graph of $$g(x)$$ lies above the $$x$$-axis and does not have any $$x$$-intercepts.

### Asymptote

Exponential functions of the form $$y = ab^{(x+p)} + q$$ have a horizontal asymptote, the line $$y = q$$.

## Example

### Question

Determine the asymptote for $$y = 5 \times 3^{(x+1)} – 1$$.

### Determine the asymptote

The asymptote of $$g(x)$$ can be calculated as: \begin{align*} 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} – 1 & \ne -1\\ \therefore y & \ne -1 \end{align*} Therefore the asymptote is the line $$y = -1$$.

### Sketching graphs of the form $$f(x)=a{b}^{(x+p)}+q$$

In order to sketch graphs of functions of the form, $$f(x)=a{b}^{(x+p)}+q$$, we need to determine five characteristics:

• shape

• $$y$$-intercept

• $$x$$-intercept

• asymptote

• domain and range

## Example

### Question

Sketch the graph of $$2y = \text{10} \times 2^{(x+1)} – 5$$.

Mark the intercept(s) and asymptote. State the domain and range of the function.

### Examine the equation of the form $$y = ab^{(x + p)} + q$$

We notice that $$a > 0$$ and $$b > 1$$, therefore the function is increasing.

### Determine the $$y$$-intercept

The $$y$$-intercept is obtained by letting $$x = 0$$: \begin{align*} 2y &= \text{10} \times 2^{(0+1)} – 5\\ &= \text{10} \times 2 – 5\\ &= \text{15}\\ \therefore y &= 7\cfrac{1}{2} \end{align*} This gives the point $$(0;7\cfrac{1}{2})$$.

### Determine the $$x$$-intercept

The $$x$$-intercept is obtained by letting $$y = 0$$: \begin{align*} 0 &= \text{10} \times 2^{(x+1)} – 5\\ 5 &= \text{10} \times 2^{(x+1)} \\ \cfrac{1}{2} &= 2^{(x+1)}\\ 2^{-1} &= 2^{(x+1)}\\ \therefore -1 &= x + 1 \quad \text{(same base)}\\ -2 &= x \end{align*} This gives the point $$(-2;0)$$.

### Determine the asymptote

The horizontal asymptote is the line $$y = -\cfrac{5}{2}$$.

### State the domain and range

Domain: $$\{ x: x \in \mathbb{R} \}$$

Range: $$\{ y: y > -\cfrac{5}{2}, y \in \mathbb{R} \}$$

## Example

### Question

Use the given graph of $$y = -2 \times 3^{(x + p)} + q$$ to determine the values of $$p$$ and $$q$$.

### Examine the equation of the form $$y = ab^{(x + p)} + q$$

From the graph we see that the function is decreasing. We also note that $$a = -2$$ and $$b = 3$$.

We need to solve for $$p$$ and $$q$$.

### Use the asymptote to determine $$q$$

The horizontal asymptote $$y = 6$$ is given, therefore we know that $$q = 6$$. $y = -2 \times 3^{(x + p)} + 6$

### Use the $$x$$-intercept to determine $$p$$

Substitute $$(2;0)$$ into the equation and solve for $$p$$: \begin{align*} y &= -2 \times 3^{(x + p)} + 6 \\ 0 &= -2 \times 3^{(2 + p)} + 6 \\ -6 &= -2 \times 3^{(2 + p)} \\ 3 &= 3^{(2 + p)} \\ \therefore 1 &= 2 + p \quad \text{(same base)}\\ \therefore p &= -1 \end{align*}

### Write the final answer

$y = -2 \times 3^{(x – 1)} + 6$